Decomposable Tensors Problem: V Vector Space Dim ≤ 3

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In summary: This means that v is not a homogeneous element, since it is not in T^1(V). Therefore, the decomposition would not be valid in this case. In summary, if a vector space V has dimension less than or equal to 3, then every homogeneous element in the exterior algebra \Lambda(V) is decomposable, meaning it can be written as a single exterior product of two forms of lower degree. However, the decomposition must involve forms of different degrees, otherwise it would not be a valid decomposition.
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Problem:
V a vector space with [itex]dimV \le 3[/itex], then every homogeneous element in [itex]\Lambda(V)[/itex] is decomposable.


So, this exercise doesn't sound very difficult. My problem is, that i don't know the definition of homogeneous and decomposable. Can you please help me?
Thank you
 
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  • #2
A homogeneous element of [tex]\Lambda(V)[/tex] is a form of definite degree, [tex]p[/tex]. A decomposable form is one that can be written as a single exterior product of two forms of lower degree [tex]\omega = \mu \wedge \nu[/tex].
 
  • #3
Hello,

Thank you for your help!
are these examples correct?:
For dimV=2 We have [itex]\Lambda(V) =[/itex] IK [itex]\times T^1(V) \times T^2(V)[/itex].

-Then a homogeneous elm. would be all elm. in [itex]v\in \IK[/itex] or [itex]v\in T^1(V)[/itex] or [itex]v\in T^2(V)[/itex].
Where T is the Alternator.

-Is it allowed to use the single exterior product of two form of equal degree, i.e. [itex]v=\mü \wedge \tau[/itex] with [itex]v,\mü,\tau \in T^1(V)[/itex] for instance?


Thanks
 
  • #4
Canavar said:
Hello,

Thank you for your help!
are these examples correct?:
For dimV=2 We have [itex]\Lambda(V) =[/itex] IK [itex]\times T^1(V) \times T^2(V)[/itex].

-Then a homogeneous elm. would be all elm. in [itex]v\in \IK[/itex] or [itex]v\in T^1(V)[/itex] or [itex]v\in T^2(V)[/itex].
Where T is the Alternator.

Yes a homogeneous element would be in [tex]T^p(V)[/tex] for some [tex]p[/tex].

-Is it allowed to use the single exterior product of two form of equal degree, i.e. [itex]v=\mü \wedge \tau[/itex] with [itex]v,\mü,\tau \in T^1(V)[/itex] for instance?

Yes it is possible that the decomposition involves forms of equal degree. However, in your example, if [tex]\mu, \tau \in T^1(V)[/tex], then [tex] v= \mu \wedge\tau\in T^2(V)[/tex].
 

Related to Decomposable Tensors Problem: V Vector Space Dim ≤ 3

1. What is the Decomposable Tensors Problem?

The Decomposable Tensors Problem is a mathematical problem that involves finding decomposable tensors in a vector space with dimension less than or equal to three. A tensor is a mathematical object that can be represented as a multi-dimensional array and has applications in physics, engineering, and other fields.

2. What is the significance of V Vector Space Dim ≤ 3 in this problem?

The condition V Vector Space Dim ≤ 3 means that the vector space in which the decomposable tensors are being searched for has a dimension of three or less. This limit allows for a more manageable and solvable problem, as the complexity of the calculations increases significantly as the dimension of the vector space increases.

3. How is this problem relevant to scientific research?

This problem has applications in various scientific fields, including physics, material science, and computer science. The solution to the Decomposable Tensors Problem can lead to a better understanding of the properties and behavior of materials, as well as improvements in data compression and machine learning algorithms.

4. What are some techniques used to solve the Decomposable Tensors Problem?

Some common techniques used to solve this problem include matrix factorization, eigenvalue decomposition, and tensor decomposition methods, such as Tucker decomposition and CANDECOMP/PARAFAC (CP) decomposition.

5. Can the Decomposable Tensors Problem be extended to higher dimensions?

Yes, the problem can be extended to higher dimensions, but the complexity of the calculations increases significantly. Additionally, the number of decomposable tensors in higher dimensions is much larger, making the problem more challenging to solve.

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